\newproblem{lay:2_2_17}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.17}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose $A$, $B$ and $C$ are invertible $n\times n$ matrices. Show that $ABC$ is also invertible by producing a matrix $D$ such that
	$(ABC)D=I=D(ABC)$
}{
  % Solution
	The sought matrix $D$ is
	\begin{center}
		$D=C^{-1}B^{-1}A^{-1}$
	\end{center}
	Let us check that this matrix is actually the inverse of $ABC$.
	\begin{center}
		$\begin{array}{rcl}(ABC)D&=&(ABC)(C^{-1}B^{-1}A^{-1})=AB(CC^{-1})B^{-1}A^{-1}\\
		   &=&ABB^{-1}A^{-1}=AA^{-1}=I
		\end{array}$\\
		$\begin{array}{rcl}D(ABC)&=&(C^{-1}B^{-1}A^{-1})(ABC)=C^{-1}B^{-1}(A^{-1}A)BC\\
		   &=&C^{-1}B^{-1}BC=C^{-1}C=I
		\end{array}$\\
	\end{center}
}
\useproblem{lay:2_2_17}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
